# Are $C^1$ mappings locally Lipschitz?

Let $$g:\mathbb{R}^n\times \mathbb{R}^{n'}\times [0,1]\rightarrow \mathbb{R}^n$$ a continuously differentiable function. Can we deduce that $$g$$ is locally Lipschitz ?

which is equivalent that for all nonempty compact sets $$K\subset\mathbb{R}^n\times \mathbb{R}^{n'}\times [0,1]$$, there exosts $$M>0$$ such that $$\| g(x_1,y_1,t)-g(x_2,y_2,t) \| \leq M(\|x_1-x_2 \| + \|y_1-y_2 \| ),$$ for all $$(x_1,y_1,t),(x_2,y_2,t) \in K$$ ?

Or should we assume some additional convexity hypothesis ?

• No, you don't need an additional convexity hypothesis. You can always enlarge $K$ to a convex compact set. Jun 22, 2021 at 8:58

Yes, we can already deduce local Lipschitzianity. Just note that $$\sup_{(x, y, t) \in \overline{B} \times J} \lVert Dg(x, y, t) \rVert < \infty$$ for some ball $$B \subseteq \mathbb{R}^n \times \mathbb{R}^{n'}$$ and interval $$J$$ such that $$B \times J \supseteq K$$ because of domain compactness and continuity of $$Dg$$. We can choose such ball and interval because $$K$$ is bounded.
Then use the Mean Value Theorem to follow Lipschitz-continuity on $$\overline{B} \times J$$ which implies Lipschitz-continuity on $$K$$.